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Fudgin [204]
2 years ago
14

Why do you think a removable discontinuity (hole) doesn't produce an asymptote on the graph of a polynomial function, even thoug

h it is excluded from the domain of the function?
Mathematics
1 answer:
Monica [59]2 years ago
8 0
Because  removable discontituity means that the limit of the function at that point has a finite value, and then you define the value of the function as that valu (the limit value).

An asymptote means that the limit of the function goes to positive or negative infinity.

You cannot meet both conditions, finite  and infinity limit at the same time.
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What is the y-intercept of the function,represented by the table of values below?
Hitman42 [59]

Answer:

8

Step-by-step explanation:

So the y-intercept is not given by your table because there is no x that is listed as 0.

But don't fret; we can still find it.

Let's see if the function is linear by seeing if we have the same slope per two points in the table.

For the first pair ( the points (-2,16) and (1,4) ), x increased by 3 and the y decreased by 12 so the slope there is -12/3=-4.

Now looking at the next pair ( the points (1,4) and (2,0) ), x increased by 1 while y decreased by 4 so the slope is -4/1=-4.

So the function appears to be linear.

So the slope-intercept form of a line is y=mx+b where m is slope and b is y-intercept.

We already found the slope from earlier which is m=-4.

So the equation so far is y=-4x+b.

Now to find b, the y-intercept, we need to use a point (x,y) on the line along with y=-4x+b.

Let's see my favorite on the list of points is (2,0).

y=-4x+b with (x,y)=(2,0)

0=-4(2)+b

0=-8+b

8=b

So the y-intercept is 8.

8 0
2 years ago
Read 2 more answers
Find the inverse of y=x2-10x
anzhelika [568]
<span>this is pretty hard but here is your answer 
</span>

y = x^2 - 10x + 25 - 25

<span> y = (x-5)^2 - 25 </span>

<span> y+25 = (x-5)^2 </span>

<span> x-5 = +/-sqrt(y+25) </span>

 

<span> And you get TWO inverses: </span>

 

<span> x = 5 + sqrt(y+25), for x>=5 </span>

<span> x = 5 - sqrt(y+25), for x<=5</span>


5 0
2 years ago
Read 2 more answers
Rewrite the expression (3a^2b^3c^5)/(8x^4y^3z)so that it has no denominator. (note: do not use a fraction line in your answer. a
Gemiola [76]
Use:

\dfrac{1}{a}=a^{-1}\\\\\dfrac{1}{a^n}=a^{-n}

\dfrac{3a^2b^3c^5}{8x^4y^3z}=3a^2b^3c^5\cdot8^{-1}x^{-4}y^{-3}z^{-1}

Answer: 3"8^-1a^2b^3c^5x^-4y^-3z^-1
5 0
2 years ago
Read 2 more answers
Match the following items by evaluating the expression for x = -6.
Kryger [21]

Step-by-step explanation:

Put the value of x = -6 to all expressions:

x^{-2}=(-6)^{-2}=\dfrac{1}{(-6)^2}=\dfrac{1}{6}\qquad\text{used}\ a^{-n}=\dfrac{1}{a^n}\\=====================\\x^{-1}=(-6)^{-1}=\dfrac{1}{(-6)^1}=-\dfrac{1}{6}\\=====================\\x^0=(-6)^0=1\qquad a^0=1\ \text{for all real numbers except 0}\\=====================\\x^1=(-6)^1=-6\qquad a^1=a\ \text{for all real numbers}\\=====================\\x^2=(-6)^2=36

5 0
2 years ago
Find a ⋅ b. a = 8i + 6j, b = 4i + 5j
frutty [35]
You could rewrite this as double brackets, as you are multiplying together two sets of two terms. It would then look like:
(8i + 6j)(4i + 5j)
and you can expand by multiplying together all of the terms

8i × 4i = 32i²
8i × 5j = 40ij
6j × 4i = 24ij
6j × 5j = 30j²

To get your final answer, you then just need to add together all of the like terms, and get 32i² + 30j² + 64ij

I hope this helps!
7 0
2 years ago
Read 2 more answers
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